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Floats on Mercury

  • Thread starter talaroue
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  • #1
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1. Homework Statement
A 17.89 cm diameter, 42.43 cm tall steel cylinder (rsteel=7900 kg/m3) floats in mercury. The axis of the cylinder is perpendicular to the surface. What length of steel is above the surface?


2. Homework Equations

Pb=Pa+density*g*h


3. The Attempt at a Solution

Pa=101300
Pb?
density= 7900 kg/m^3
H=(total length-submerged)
g=(9.8)

Pb=101300+((9.8*7900*(total length-submerged))

I have 2 unkowns. What would Pb be?
 

Answers and Replies

  • #2
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Any help would be much apperiated!
 
  • #3
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For some reason I am having a hard time grasping pressure.
 
  • #4
Redbelly98
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It looks like you're using Pb for the pressure on the bottom face. You can express that in terms of the density of mercury and the depth.
 
  • #5
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Will Pb(pressure on bottom) always equal the denisty * depth? That is the density of the force acting against the object in this case......mercury?
 
  • #6
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so i have RHO(mercury)*H(submerged)=101300(Pa)+(RHO*G*LENGTH OF CYLINDER)

solve for h i have (101300+(7900*9.8*.4243))/13600= 9.8 m.....?
 
  • #7
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i tried 0 seeing if it was a trick question but the trick was on me because it was wrong
 
  • #8
Redbelly98
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Will Pb(pressure on bottom) always equal the denisty * depth? That is the density of the force acting against the object in this case......mercury?
Almost. It's actually ρmercury·g·hsubmerged, plus the pressure at the surface of the mercury.
 
  • #9
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So Pb= pressure on surface of mercury+denisty *g*h? that doesn't seem right
 
  • #10
Redbelly98
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What about it does not seem right to you?

Things about it that are right (if you think about it):

  • Pressure at the surface (h=0) is the pressure at the surface, i.e. atmospheric pressure.
  • Pressure increases with depth h
  • Pressure is greater for denser fluids
  • Pressure is greater for greater values of g
 
  • #11
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What doesn't seem right is that Pb is pressure at bottom, so why would i need the pressure at the top.
Pb=Pa+rho*g*h
^ ^
^ ^
^ Pressure at above which in this case is atmosphereic pressure
^
Pressure at bottom

So if the pressure at the bottom is just (density of mercury*h*g) that works right? Where h is the amount submerged

The pressure at the top/above is jsut 101300 Pa

rho is denisty of the stell which is given 7900, g is 9.8 and h is the total height

Where am I going wrong?
 
  • #12
Redbelly98
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So if the pressure at the bottom is just (density of mercury*h*g) that works right? Where h is the amount submerged
No, that doesn't work. What happens when h is zero? According to you, the pressure would be zero. But we know that the pressure actually equals atmospheric pressure at h=0.
 
  • #13
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Isn't that why we have pressure above the can?
 
  • #14
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Is there another way to appoarch this?
 
  • #15
Redbelly98
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Last edited by a moderator:
  • #16
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I can use the density of steel to see how much steel is present, then this is also equal to the weight of displaced mercury. CORRECT?

Then see how deep it is using m=Vp. CORRECT?
 
  • #17
Redbelly98
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Yes.
 

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